A cell can deposit atomic mass Al/3 (about 27/3 = approx 9g) Al when 96,485 coulombs is passed through a solution. How many coulombs do you have? That's coulombs = amperes (that's amps) x seconds.
So you have C = 5.00 x 5.00 hrs x (60 min/1 hr) x (60 sec/1 min) = about 9E4.
The problem boils down to this. 96,485 C will deposit 9g Al; how much Al will be deposited by 9E4 C.
That's 9g x (9E4/96,485) = ?
That 9 is a close estimate. You should recalculate these numbers.
2.
I don't know how Ecell can be calculated without knowing ??M concn Cu in the right hand cell. I suspect you meant Eocell.
B. If Ecell is 0.084, then
Ecell = Eocell - (0.0592/n)log (dilute)/(concd) and solve for dilute.
How would you start and finish these questions by a step by step process? Does Amp equal to Joules?
1. How many grams of aluminum can be produced by the electrolysis of aqueous aluminum chloride by a current of 5.00 amps for 5.00 hours?
2. Consider the following electrochemical cell: Cu(s) | Cu^2+(aq, 0.10 M) || Cu^2+(aq), ?? M | Cu(s)
A) What is the overall cell reaction and Ecell?
B) Determine the unknown [Cu^2+] if Ecell = 0.084 V at 25°C.
1 answer
1 answer